Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.11 - Hyperbolic Functions - 3.11 Exercises: 19

Answer

$e^{nx}$

Work Step by Step

$(coshx+sinhx)^{n}$ = cosh nx + sinh nx Convert the left side to the $e^{x}$ forms and simplify $(\frac{e^{x} + e^{-x}}{2} + \frac{e^{x} - e^{-x}}{2} )^{n}$ = $(\frac{2e^{x}}{2})^{n}$ = $(e^{x})^{n}$ = $e^{nx}$
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